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layout="Title" style="Title">Graphical analysis of the chemostat equations</Text-field><Text-field layout="Author" style="Author">Prof. Matt Miller, Department of Mathematics, University of South Carolina 
miller@math.sc.edu 
</Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">
</Font></Text-field></Input></Group><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">Introduction</Text-field></Title><Group><Input><Text-field layout="Normal" style="Text">This worksheet explores the graphical analysis of the chemostat equations, using analytic and visualization methods to solve and explore the problem, resulting in graphs of the phase portrait, phase dynamics and chemostat dynamics of the model.</Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">Initialization</Text-field></Title><Group><Input><Text-field prompt="&gt; " style="Maple Input">restart:
interface( warnlevel=0 ):
with(plots):
with(DEtools):</Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">Create Initial Equations</Text-field></Title><Group><Input><Text-field prompt="&gt; " style="Maple Input">eqn18a:=diff(N(t),t)=alpha1*(C(t)/(1+C(t)))*N (t)-N(t);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">eqn18b:=diff(C(t),t)=-(C(t)/(1+C(t)))*N(t) -C(t) + alpha2;</Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1"><Font executable="false">The nullclines</Font></Text-field></Title><Section collapsed="true"><Title><Text-field layout="Heading 2" style="Heading 2">The N nullcline - solutions of the dN/dt equation</Text-field></Title><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">The growth rate of the population (dN/dt) = 0 under two possible scenarios.  We can find the two scenarios by setting the right hand side of the equation equal to zero and solving for N or C.</Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">The first solution is the value of N for which dN/dt = 0</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">N_value_1:=solve(rhs(eqn18a)=0,N(t));</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">The second solution if the value of C for which dN/dt=0</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">C_value_1:=solve(rhs(eqn18a)=0,C(t));</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">This solution C=1/(</Font><Equation input-equation="alpha1;" style="2D Comment">NiMlJ2FscGhhMUc=</Equation><Font background="[0,0,0]" family="Times New Roman"> -1) is a line parallel to the N axis, because the value of C does not depend upon N.  Its position on the C axis is at 1/(</Font><Equation input-equation="alpha1;" style="2D Comment">NiMlJ2FscGhhMUc=</Equation><Font background="[0,0,0]" family="Times New Roman"> -1).</Font></Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 2" style="Heading 2">The C nullcline - solutions of the dC/dt </Text-field></Title><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">One solution is the value of N for which dC/dt=0</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">N_value_2:=solve(rhs(eqn18b)=0,N(t));</Text-field></Input></Group><Group><Input><Text-field layout="Normal" prompt="&gt; " style="Maple Input">N_value_2 := factor(N_value_2);</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">This solution, N=(</Font><Equation input-equation="alpha2;" style="2D Comment">NiMlJ2FscGhhMkc=</Equation><Font background="[0,0,0]" family="Times New Roman"> - C)(1+C)/C is a curve whose value is N=0 when C=</Font><Equation input-equation="alpha2" style="2D Comment">NiMlJ2FscGhhMkc=</Equation></Text-field></Input></Group></Section></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">The Model</Text-field></Title><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">Since the chemostat model involves two equations in two variables, we can plot the behavior of the model on a graph whose axes are N and C.  It is an arbitrary decision, which variable goes on which axis.  This kind of plot is called a <Font background="[0,0,0]" family="Times New Roman">phase portrait<Font background="[0,0,0]" family="Times New Roman"> or <Font background="[0,0,0]" family="Times New Roman">phase-plane portrait<Font background="[0,0,0]" family="Times New Roman">.  Edelstein-Keshet plots her chemostat model with N as the horizontal axis and C as the vertical axis.  If  we want to plot the solutions of the N equation and the C equation on these axes, we need to define the solutions as functions of N: </Font></Font></Font></Font></Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">for instance  C=f(N) for the solution of the dN/dt=0 equation </Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">and   C=g(N) for the solution of the dC/dt=0 equation.</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">f:=C_value_1;</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">Finding g(N) is a bit more complicated, because we have an equation:</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">N=N_value_2;</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input"/></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">Unfortunately it is of the form N=some function of C rather than C = some function of N, so we need to rearrange it to put C on the left hand side and something different on the right hand side.  To do this, we solve this equation for C and see what happens</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">g:=solve(N(t)=N_value_2,C(t));</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">This solution is quite messy, and actually is a pair of solutions - note the comma after the first square root.  We can see what these two solutions look like when plotted on the axes C and N.  To do so, we will define g1(N) to be the first solution and g2(N) as the second:</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">g1:=op(1,[g]);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">g2:=op(2,[g]);</Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">Plotting the Model</Text-field></Title><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">If we want to plot the graph, we need to define some specific values of alpha1 and alpha2.  Looking at the equations, it is clear that alpha1 must be greater than 1, or else the equilibrium nutrient concentration will be zero or negative. Alpha2 must be positive.</Font></Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">The following is the statement that defines the values of alpha1 and alpha2 in preparation for plotting graphs of the phase-portrait</Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">Our starting parameter values are alpha1=4 and alpha2=1</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">parameters:=alpha1=4, alpha2=1;</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">G:=subs(parameters,g1);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">F:=subs(parameters,f);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">C_nullcline:=plot(G,N=0..20,color=red):</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">N_nullcline:=plot(F,N=0..20,color=green):</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">display(C_nullcline,N_nullcline);</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">The alternate function g(N) is g2, and its graph is</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">G:=subs(parameters,g2);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">C_nullcline1:=plot(G,N=0..20,color=blue):</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">display(C_nullcline1,N_nullcline);</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">The only <Font background="[0,0,0]" family="Times New Roman">feasible<Font background="[0,0,0]" family="Times New Roman"> solution is the one with positive values of the nutrient concentration and positive values of the bacterial population.</Font></Font></Font></Text-field></Input></Group></Section><Section collapsed="true"><Title><Text-field layout="Heading 1" style="Heading 1">Changing the Model</Text-field></Title><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">Try changing the values of the parameter alpha1 and see how that changes the positions of the equilibrium lines.  Remember that for a feasible value of C = 1/(alpha1-1), it is necessary that </Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">alpha1&gt;1, otherwise C would be infinite (alpha1=1) or negative (alpha1 &lt; 1).  </Font></Text-field><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">See what happens if you violate these criteria.</Font></Text-field></Input></Group><Group><Input><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">Edelstein-Keshet notes another constraint on the model in expression 21 on p 194.  She claims that </Font></Text-field><Text-field><Font background="[0,0,0]" family="Times New Roman">alpha2 &gt; 1/(alpha1-1) must be true.  What happens if this is not true?  Try it and see, by changing the value of alpha2  in the parameters statement above.</Font></Text-field></Input></Group><Group><Input><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">We calculated an overall equilibrium for the chemostat system (see also equations 20-25 on pp 128-129 in Edelstein-Keshet.  Where is that equilibrium on the phase portrait?</Font></Text-field></Input></Group><Group><Input><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">If the nullclines are lines defining the conditions dN/dt=0 (N nullcline) and dC/dt=0 (C nullcline), presumably on one side of these lines the growth rates are positive and on the other side the growth rates are negative.  </Font></Text-field><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">How would you decide which side of the line has positive dN/dt or dC/dt and which side has negative values?</Font></Text-field></Input></Group><Group><Input><Text-field/><Text-field><Font background="[0,0,0]" family="Times New Roman">We will now try looking at the dynamics of the chemostat system, using your parameters.  In order to do this, we define the equations, the variables, the time domain, the initial conditions, and tell the system to plot the lines in particular colors.  This is done with the DEtools package and the DEplot command.</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">equations:=subs(parameters,[eqn18a,eqn18b]);</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">vars:=[N(t),C(t)];</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">domain:=t=0..20;</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">initial_conditions:=[[0,1,0],[0,3,1],[0,0.01,0.01],[0,5,.01]];</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">phase_dynamics:=DEplot(equations,vars,domain,initial_conditions,scene=[N,C],linecolor=[red,green,blue,black],stepsize=0.1):</Text-field></Input></Group><Group><Input><Text-field/></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">display(phase_dynamics);</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">We can now display the dynamics on top of the nullcline graphs</Font></Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">display(phase_dynamics,N_nullcline,C_nullcline);</Text-field></Input></Group><Group><Input><Text-field/></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">C_dynamics:=DEplot(equations,vars,domain,initial_conditions,scene=[t,C],linecolor=[red,green,blue,black],stepsize=0.1):</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">N_dynamics:=DEplot(equations,vars,domain,initial_conditions,scene=[t,N],linecolor=[pink,aquamarine,navy,grey],stepsize=0.1):</Text-field></Input></Group><Group><Input><Text-field prompt="&gt; " style="Maple Input">display(C_dynamics,N_dynamics,title="Chemostat Dynamics - (C dark, N pale)");</Text-field></Input></Group><Group><Input><Text-field><Font background="[0,0,0]" family="Times New Roman">This is the dimensionless version of the model.  What is the scale of the nutrient values?  What does it mean when nutrients are less than 1 or greater than 1?  What is the scale of population?</Font></Text-field></Input></Group></Section><Group><Input><Text-field layout="Normal" style="Text"><Font italic="true">
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